Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Why can't I solve this "easy" question?

  • 07-03-2016 9:02pm
    #1
    Dughorm
    Closed Accounts Posts: 895 ✭✭✭


    If x+y=5 and x^y = 8 find x and y? :eek:

    It's lack of a method rather than the actual answer that bothers me.


Comments

  • #2
    ted1
    Registered Users, Registered Users 2 Posts: 23,900 ✭✭✭✭ted1


    Dughorm wrote: »
    If x+y=5 and x^y = 8 find x and y? :eek:

    It's lack of a method rather than the actual answer that bothers me.

    Get x or y on its own. Y= 5-x
    Then replace x in the next one.
    X^(5-x)=8
    X^5-x^2=8
    X^3=8
    X=2
    Y=3


  • #3
    cruizer101
    Registered Users, Registered Users 2 Posts: 3,205 ✭✭✭cruizer101


    How do you do this:
    X^(5-x)=8
    X^5-x^2=8

    Where does the two come from, is x^(5-x) not equal to (x^5)/(x^-x).

    I would of thought needed logs


  • #4
    GarIT
    Registered Users, Registered Users 2 Posts: 8,671 ✭✭✭GarIT


    cruizer101 wrote: »
    How do you do this:
    X^(5-x)=8
    X^5-x^2=8

    Where does the two come from, is x^(5-x) not equal to (x^5)/(x^-x).

    I would of thought needed logs

    x^(5-x) = (x^5) - (x^2) it makes more sense when spread out but I'm thinking it should be (x^5) - (x^x) instead


  • #5
    Dughorm
    Closed Accounts Posts: 895 ✭✭✭Dughorm


    ted1 wrote: »
    Get x or y on its own. Y= 5-x
    Then replace x in the next one.
    X^(5-x)=8
    X^5-x^2=8
    X^3=8
    X=2
    Y=3

    So we're talking

    X^(5-x) = 8
    X^5-X^X = 8
    logX^5 - log X^X = log 8
    5logX - Xlog X = log 8

    Where does it go from here?


  • #6
    ZorbaTehZ
    Registered Users, Registered Users 2 Posts: 2,149 ✭✭✭ZorbaTehZ


    Almost every post in this thread so far has a false statement(s) or deduction(s) in it! E.g. x^(5-x) is not equal to x^5-x^2, just try x=1 or x=2 to see why!

    Regarding the original question, the type of equations you posted are transcendental in general and will not have a closed form solution. In your case you have a solution in integers just by inspection.


  • Advertisement
Advertisement